Question #494a3

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Answer 1 · 2015-01-03T23:47:47

A photon of wavelength 280nm will deliver this amount of energy.

#E=hf#

#E#= energy in #J#

#f# = frequency in #s^(-1)#

#h# = Planck's Constant = #6.63xx10^(-34)Js#

#f=E/h=(7.12xx10^(-19))/(6.63xx10^(-34))=1.074xx10^(15)s^(-1)#

#c=flambda#

#lambda=(3xx10^8)/(1.074xx10^(15))=2.8xx10^(-7)m =280nm#

This occurs in the uv part of the spectrum. You could use a uv lamp or a tuneable laser.